I think the problem here is that you haven't stated the complete question and Sam guessed at it. The question Sam answered is:
If you choose 6 melons from one of the stores what is the probability that the total weight from Store A would be greater than what you would have gotten if you selected 6 from store B.
But your previous question asked
If you choose 3 melons from either store A or store B which selection has the higher probability that the total weight is greater than 8 kg?
Is it this second question with 6 substituted for 3 and possibly a different weight replacing 8?
If it is the latter than just as your instructors did for the previous question you need to compute the 2 z scores for the sums of 6 iid melon weights from store A and store B. Use the standard deviations and means as you correctly described them above in your question and compare the probabilities to 8 kg or whatever value you need to substitute for it.
Based on your comment Sam was giving you the right approach. Let me just fill in somemore detail so that you can finish it off yourself. As he pointed out
if X and Y are two independent normal ranom variance the sum X-Y is normal and has mean equal to the difference of the two means and variance equal to the sum of the two variances. So in this probem Take W$_A$-W$_B$ and divide it by sqrt(6*0.7^2+6*0.2^2) to get the Standard normal random variable. Then the z score is
[μA - μB)]/sqrt(6*0.7^2+6*0.2^2). So look at the standard normal table to see what the probability is that a standard normal is larger than this z score. (In all these problem if the sample standard deviation is used t distribution should be used. But in this case since the variances are
different the Welch test would be used instead of the t test that pools the variance estimates.)