Answer
$V_{rms} = 22.4~V$
Work Step by Step
We can write an expression for the power:
$P = \frac{V^2}{R}$
Note that: $~~P \propto V^2 \propto V_{rms}^2$
If the power dissipated increases by a factor of 5, then $V_{rms}$ increases by a factor of $\sqrt{5}$
We can find the required value for the $V_{rms}$:
$V_{rms} = (\sqrt{5})~(10.0~V) = 22.4~V$
