b
(B)
Heat current will flow radially outward.
We know $R _{ T }=\frac{\ell}{ kA }$
Here $\int dR _{ T }=\int \limits_{ R _1}^{ R _2} \frac{ dx }{ k 2 \pi x \ell}$
$\Rightarrow R _{ T }=\frac{1}{2 \pi k \ell} \ln \left(\frac{ R _2}{ R _1}\right)$
$\frac{ dQ }{ dt }=\frac{\Delta T }{ R _{ T }}$
$\therefore \frac{ dQ }{ dt }=\frac{(110-10) 2 \pi(0.38)(10)}{\ell n(2)}$
$=3444.6 \approx 3445 \,kW$
