Answer
682.5 K = 409.5 Degree Celsius
Work Step by Step
Let $T_{1} = 0^{o}C + 273 K = 273 K$, $V_{1} = 4 L$, $V_{2} = 10 L$, $T_{2} = ?$
Using Charles's Law, we have:
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{4 L}{273 K} = \frac{10 L}{T_{2}}$
$T_{2} = \frac{10 L}{4 L}$ x 273K
= 682.5 K