Answer
$ 4 \ \mathrm{times \ brighter} $
Work Step by Step
The relation can be expressed as $ \displaystyle I = \frac{k}{d^2} $.
When the chair is halfway, the distance is $ \displaystyle \frac{1}{2}d $:
$$ \begin{align*}
\displaystyle I &= \frac{k}{d^2} \\\\
I &= \frac{k}{\left(\frac{1}{2}d\right)^2} \\\\
I &= \frac{k}{\frac{d^2}{4}} \\\\
I &=4\left( \frac{k}{d^2} \right)\\\\
\end{align*} $$
So, the original brightness will be multiplied by $ 4$.
