1. Reaction involved

Ethylamine is a primary aliphatic amine: $\mathrm{C_2H_5NH_2}$

Primary aliphatic amines react with nitrous acid generated from $\mathrm{NaNO_2/HCl}$ to give alcohol with evolution of nitrogen gas:

$\mathrm{C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O}$

So, 1 mole of ethylamine gives 1 mole of $\mathrm{N_2}$.

2. Moles of nitrogen gas evolved

Given volume of $\mathrm{N_2}$ at STP is $2.24\,\mathrm{L}$.

At STP, $1\text{ mole gas} = 22.4\,\mathrm{L}$

Hence, $n(\mathrm{N_2})=\frac{2.24}{22.4}=0.1\text{ mol}$

Therefore, $n(\mathrm{ethylamine})=0.1\text{ mol}$

3. Molar mass of ethylamine

Ethylamine: $\mathrm{C_2H_7N}$

$M = 2(12) + 7(1) + 14 = 24 + 7 + 14 = 45\,\mathrm{g\,mol^{-1}}$

4. Mass of ethylamine

$m = n \times M = 0.1 \times 45 = 4.5\,\mathrm{g}$

Now the question asks:

$X = \underline{\hspace{1cm}} \times 10^{-1}\,\mathrm{g}$

Since $4.5\,\mathrm{g} = 45 \times 10^{-1}\,\mathrm{g}$

Therefore, $\boxed{X=45}$