Answer
$\theta_{min}=\cos^{-1}\Big(1-\frac{1}{4.98\times10^{74}}\Big)$
Work Step by Step
$\sqrt {l(l+1)}\hbar\cos\theta=m_l\hbar$
or, $\cos\theta=\frac{m_l}{\sqrt {l(l+1)}}$
or, $\theta=\cos^{-1}\Big(\frac{m_l}{\sqrt {l(l+1)}}\Big)$
For $\theta_{min}$, the value of $m_l$ will be maximum and here it is $m_l=l$ Thus,
$\theta=\cos^{-1}\Big(m_l\times\frac{1}{l}(1-\frac{1}{2l})\Big)$
Putting the given values
$\theta_{min}=\cos^{-1}\Big(l\times\frac{1}{l}(1-\frac{1}{2l})\Big)$
or, $\theta_{min}=\cos^{-1}\Big(1-\frac{1}{2l}\Big)$
Here, $l=2.49\times10^{74}$, Thus
$\theta_{min}=\cos^{-1}\Big(1-\frac{1}{2\times 2.49\times10^{74}}\Big)$
or, $\theta_{min}=\cos^{-1}\Big(1-\frac{1}{4.98\times10^{74}}\Big)$
