Answer
$$3\bar{0}0\ V^2/\Omega$$
Work Step by Step
We have
$$
\frac{(120\ V)^2 }{ 47.6\ \Omega}=302.52\ V^2/\Omega
.$$
Now, round this product to two significant digits, which is the accuracy of the least accurate measurement. That is,
$$302.52\ V^2/\Omega=3\bar{0}0\ V^2/\Omega$$
