1. Given data

• Monobasic strong acid
• Molar mass of acid $= 24.2\,\text{g mol}^{-1}$
• Density of acid solution $= 1.21\,\text{kg L}^{-1} = 1210\,\text{g L}^{-1}$
• Volume of NaOH $= 25\,\text{mL} = 0.025\,\text{L}$
• Molarity of NaOH $= 0.24\,\text{M}$

2. Find moles of NaOH

Since, $n = M \times V$

So, $n(\text{NaOH}) = 0.24 \times 0.025 = 0.006\,\text{mol}$

3. Neutralization stoichiometry

The acid is monobasic, so it provides one $\text{H}^+$ per mole.

Neutralization reaction: $\text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O}$

Thus, molar ratio is $1:1$.

Hence, required moles of acid: $n(\text{acid}) = 0.006\,\text{mol}$

4. Find molarity of the acid from density and molar mass

Assuming the given strong acid is taken as pure liquid acid, mass of $1\,\text{L}$ acid is: $1210\,\text{g}$

Number of moles in $1\,\text{L}$: $M = \frac{1210}{24.2} = 50\,\text{mol L}^{-1}$

So, molarity of acid $= 50\,\text{M}$.

5. Find volume of acid required

Using, $V = \frac{n}{M}$

$V = \frac{0.006}{50} = 1.2 \times 10^{-4}\,\text{L}$

Convert to mL: $1.2 \times 10^{-4} \times 1000 = 0.12\,\text{mL}$

6. Express in the required form

We need: $\underline{\hspace{1cm}} \times 10^{-2}\,\text{mL}$

Now, $0.12\,\text{mL} = 12 \times 10^{-2}\,\text{mL}$

So the required integer is: $\boxed{12}$