A small object is placed at the centre of a large evacuated hollow...

Using Stefan-Boltzmann law,

$P = \dfrac{dQ}{dt} = \sigmaeA\left[{T}_{B}^{4} - {T}_{S}^{4}\right], \left(∵ {T}_{S} = 0K\right)$

So,

$\dfrac{dQ}{dt} = e\sigmaA{T}^{4}, \left(\text{consider}, T = {T}_{B}\right)$

also,

$\dfrac{dQ}{dt} = - ms\dfrac{dT}{dt} = e\sigmaA{T}^{4}$

$\mathop{- ∫}_{200}^{T}\dfrac{dT}{{T}^{4}} = {∫}_{0}^{t}kdt, \text{where}k = \dfrac{\sigmaA}{ms}$

$⇒ \dfrac{1}{{T}^{3}} - \dfrac{1}{{\left(200\right)}^{3}} = 3kt$

Now,

For case \left(1\right)$</p><p>$\dfrac{1}{{\left(100\right)}^{3}} - \dfrac{1}{{\left(200\right)}^{3}} = 3k{t}_{1} . . . $\left(1\right)$

For case $\left(2\right)$

$\dfrac{1}{{\left(50\right)}^{3}} - \dfrac{1}{{\left(200\right)}^{3}} = 3k{t}_{2} . . . \left(2\right)$

Divide equation \left(2\right)$by \left(1\right)$,

$\dfrac{\dfrac{1}{{\left(50\right)}^{3}} - \dfrac{1}{{\left(200\right)}^{3}}}{\dfrac{1}{{\left(100\right)}^{3}} - \dfrac{1}{{\left(200\right)}^{3}}} = \dfrac{{t}_{2}}{{t}_{1}}$

So, $\dfrac{{t}_{2}}{{t}_{1}} = 9$