Explanation

(c) According to Kepler’s third law
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto R^{3}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{2}}{T_{1}} = \left( \frac{R_{2}}{R_{1}} \right)^{3/2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{2}}{T_{1}} = \left( \frac{3R}{R} \right)^{3/2}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{2}}{T_{1}} = \sqrt{27}$
$\therefore\ \ \ T_{2} = \sqrt{27}T_{1} = \sqrt{27} \times 4 = 4\sqrt{27}h$