Answer is (2)
$V_{\text {escane }}=11200 \mathrm{~m} / \mathrm{s}$
Say at temperature $T$ it attains $V_{\text {escape }}$
So, $\sqrt{\frac{3 k_{B} T}{m_{O_{2}}}}=11200 \mathrm{~m} / \mathrm{s}$
On solving we get,
$\mathrm{T}=8.360 \times 10^{4} \mathrm{~K}$