Correct Answer: $5\left (3+\sqrt{3} \right)$ m

Solution :
Let the height of pillar = $AB$ = $h$ m
Given: $CD=10$ m $\angle ACB = 45^{\circ} \quad \angle ADB = 60^{\circ}$
Let $BD = x$ m
From $\Delta ABC$
$\tan 45^{\circ} = \frac{AB}{BC}$
$\Rightarrow 1=\frac{h}{x + 10}$
$\Rightarrow h = (x + 10)$ m . . . . . . .$(i)$
From $\Delta ABD$
$\tan 60^{\circ} = \frac{AB}{BD}$
$\Rightarrow\sqrt{3} = \frac{h}{x}$
$\Rightarrow x =\frac{h}{\sqrt{3}}$ m . . . . .$(ii)$
From equation $(i)$
$\Rightarrow h = \frac{h}{\sqrt{3}} + 10$
$\Rightarrow h - \frac{h}{\sqrt{3}} = 10$
$\Rightarrow h\left(\sqrt{3} - 1\right)=10\sqrt{3}$
$\Rightarrow h = \frac{10\sqrt{3}}{\sqrt{3}-1}$
$\Rightarrow h = \frac{10\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} $
$\Rightarrow h = \frac{10\sqrt{3}(\sqrt{3} + 1)}{3 - 1}$
$\Rightarrow h = 5(3 + \sqrt{3})$ m
Hence, the correct answer is $5(3 + \sqrt{3})$ m.