Test whether we can infer that the population means differ 
Naturally, this is very insignificant and we fail to reject the null hypothesis. Is this the right calculation? 
 A: You mistakenly took ${\left(s_1^2\right)}^2$ and ${\left(s_2^2\right)}^2$ which gave you a much larger unpooled standard deviation than what it actually is.
For unequal variances, we should use Welch's Test. 
Let $$H_0 : \mu_1 = \mu_2$$
$$H_a : \mu_1 \neq \mu_2$$
We have
$$\frac{\bar{x_1}-\bar{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\sim t_v$$
where
$$\begin{align*}
v=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s_1^2}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s_2^2}{n_2}\right)^2}
&=\frac{\left(\frac{729}{81}+\frac{350}{50}\right)^2}{\frac{1}{81-1}\left(\frac{729}{81}\right)^2+\frac{1}{50-1}\left(\frac{350}{50}\right)^2}\\
&= 127
\end{align*}$$
Thus,
$$\frac{29-23}{\sqrt{\frac{729}{81}+\frac{350}{50}}}=1.5\sim t_{127}$$
But $t_{127,.025}\approx 1.97$
so we fail to reject the null hypothesis that the means differ.
Note that the p-value associated with $t_{127,.025}\approx 1.97$ is $0.136$ so we are closer to rejection than your calculations.
If we use the given value of $df=120$ we should come to a similar conclusion:
$$\frac{29-23}{\sqrt{\frac{729}{81}+\frac{350}{50}}}=1.5\sim t_{120}$$
and $t_{120,.025}\approx 1.9799$ so we fail to reject in this case as well.
A: The sample means $\bar X_1 = 29$ and $\bar X_2 = 23$ differ. The
question is whether this difference in sample means is good
evidence of a difference in population means.
Because of sampling error, sample mean may not be entirely reliable
estimates of population means. The degree of reliability depends
on population variability and sample size.
It is appropriate to use the Welch two-sample t test unless one has
background information (before seeing the current data) that population
variances may differ. If population variances are not the same, the
traditional pooled t statistic may not have a t distribution. The
discrepancy may be especially severe if (as here) sample sizes differ and the
the smaller sample comes from what may be the population with the
larger variance.
The formula in @Remy's answer is correct. In order to check computations
I put the summary statistics (sample sizes, means, and SDs) into Minitab
software, with the results shown below:
Two-Sample T-Test and CI 

Sample   N  Mean  StDev  SE Mean
1       81  29.0   27.0      3.0
2       50  23.0   18.7      2.6

Difference = μ (1) - μ (2)
Estimate for difference:  6.00
95% CI for difference:  (-1.91, 13.91)
T-Test of difference = 0 (vs ≠): 
   T-Value = 1.50  P-Value = 0.136  DF = 127

The P-value of the test is 0.136, which is considerable greater than 5%,
so we do not have evidence to reject $H_0: \mu_1 = \mu_2$ against its
two-sided alternative. 
The critical value for a two-sided test at the 5% level for DF = 127
is 1.9788. Because $|T| = 1.50 < 1.9788,$ we cannot reject at the 5% level.
[For DF 120, suggested in the problem, the critical value is 1.9799, which
leads to the same conclusion.]
Another indication that there is no significant difference between
population means is that the 95% CI for $\mu_1 - \mu_2$ (in the Minitab printout) contains $0.$
Thus 'no difference in population means' is a reasonable conclusion.
Finally, as a rough rule of thumb for a two-sample t test with both sample sizes
above 15, one can say that sample means must differ by more than about three
standard errors in order for the difference to be significant. Here, the
standard errors are $S_1/\sqrt{81} \approx 3$ and $S_2/\sqrt{50} \approx 2.6,$
whereas the absolute difference between sample means is only $|29 - 23| = 6.$ So it
is not a surprise that the precise computation of the Welch test does
not lead to rejection of $H_0.$
