Condition for maximum current in a series LCR circuit

Maximum current is drawn at resonance, so $X_L = X_C$ $\omega L = \frac{1}{\omega C}$

Therefore, $\omega = \frac{1}{\sqrt{LC}}$ and $f = \frac{1}{2\pi\sqrt{LC}}$

Given values

$L = 100\,\text{mH} = 0.1\,\text{H}$ $C = 2.5\,\text{nF} = 2.5 \times 10^{-9}\,\text{F}$

Resistance is given as $R = 200\,\Omega$, but note that resonance frequency does not depend on $R$.

Calculate $LC$

$LC = 0.1 \times 2.5 \times 10^{-9}$ $LC = 2.5 \times 10^{-10}$

Calculate $\sqrt{LC}$

$\sqrt{LC} = \sqrt{2.5 \times 10^{-10}}$ $= \sqrt{2.5} \times 10^{-5}$

Since given $a^2 = 10$, we interpret $a = \sqrt{10}$. Also, $2.5 = \frac{10}{4}$ so $\sqrt{2.5} = \frac{\sqrt{10}}{2}$

Hence, $\sqrt{LC} = \frac{\sqrt{10}}{2} \times 10^{-5}$

Now calculate frequency

$f = \frac{1}{2\pi\sqrt{LC}}$ $= \frac{1}{2\pi \left(\frac{\sqrt{10}}{2} \times 10^{-5}\right)}$ $= \frac{1}{\pi \sqrt{10} \times 10^{-5}}$ $= \frac{10^5}{\pi\sqrt{10}}$

Using the common JEE approximation $\pi \approx \sqrt{10}$, $f = \frac{10^5}{\sqrt{10}\cdot \sqrt{10}} = \frac{10^5}{10} = 10^4\,\text{Hz}$

Write in the asked form

$f = 10 \times 10^3\,\text{Hz}$

So the required integer is $\boxed{10}$