The sample sizes are known to be different, so there is no direct issue
of statistical significance as to difference in sample sizes. However, it
does make sense to wonder whether the sample sizes you mention are too small
reliably to detect a difference between the two failure rates you mention.
If your are testing that the two failure rates are the same
$H_0: \theta_1 = \theta_2$ against the alternative hypothesis
$H_a: \theta_1 \ne \theta_2$ or $H_a: \theta_1 < \theta_2,$
then may want to know what sample size $n$ is necessary
to detect a difference as great as $\delta = \theta_2 - \theta_1 = .17 - .11 = .06$
with power $.90.$ Power is the probability of rejecting $H_0$ when $\delta = .06.$
The answer depends on the significance level $\alpha$ (often 5%) you use for
the test. It would also be different, depending on whether you use
a one- or two-sided alternative.
The following output from Minitab statistical software indicates that,
for a two-sided 5% level test, having power 90% would require $n_1 = n_2 \approx 700.$
So you are correct to suspect that sample sizes $n_1 = 90$ and $n_2 = 270$
are not large enough for most practical purposes.
Test for Two Proportions
Testing comparison p = baseline p (versus ≠)
Calculating power for baseline p = 0.11
α = 0.05
Comparison p Size Power Actual Power
0.17 701 0.9 0.900102
The sample size is for each group.
In the figure below, the dotted line is for 90% power $(n = 701)$ and the
solid line for 80% power $(n=524).$