$I = \dfrac{k\sin \theta}{d^2}$
From the figure:
$\cos \theta = \dfrac{a}{d}$
$d = \dfrac{a}{\cos \theta}$
$I = \dfrac{k\sin \theta}{\left( \dfrac{a}{\cos \theta} \right)^2}$
$I = \dfrac{k\sin \theta}{\dfrac{a^2}{\cos^2 \theta}}$
$I = \dfrac{k}{a^2}\cos^2 \theta \, \sin \theta$
$I = \dfrac{k}{a^2}(1 - \sin^2 \theta) \sin \theta$
$I = \dfrac{k}{a^2}(\sin \theta - \sin^3 \theta)$
$\dfrac{dI}{d\theta} = \dfrac{k}{a^2}(\cos \theta - 3\sin^2 \theta \, \cos \theta) = 0$
$\cos \theta - 3\sin^2 \theta \, \cos \theta = 0$
$1 - 3\sin^2 \theta = 0$
$\sin^2 \theta = 1/3$
$\sin \theta = 1/\sqrt{3}$
$\tan \theta = \dfrac{h}{a}$
$h = a\tan \theta$
$h = a\left( \frac{1}{\sqrt{2}} \right)$
$h = \frac{1}{\sqrt{2}}\,\,a = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\,\,a$
$h = \frac{1}{2} \sqrt{2}\,a$ answer
