1. Write the reaction

Aniline reacts with acetic anhydride to give acetanilide:

$\mathrm{C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH}$

So, the mole ratio of aniline to acetanilide is:

$1:1$

2. Find molar mass of aniline

Aniline has formula $\mathrm{C_6H_5NH_2}$, i.e. $\mathrm{C_6H_7N}$.

$M(\text{aniline}) = 6(12) + 7(1) + 14 = 72 + 7 + 14 = 93\ \mathrm{g\,mol^{-1}}$

3. Calculate moles of aniline

Given mass of aniline $= 9.3\ \mathrm{g}$

$n(\text{aniline}) = \frac{9.3}{93} = 0.1\ \mathrm{mol}$

4. Find molar mass of acetanilide

Acetanilide has formula $\mathrm{C_6H_5NHCOCH_3}$, i.e. $\mathrm{C_8H_9NO}$.

$M(\text{acetanilide}) = 8(12) + 9(1) + 14 + 16 = 96 + 9 + 14 + 16 = 135\ \mathrm{g\,mol^{-1}}$

5. Calculate mass of acetanilide formed

Since mole ratio is $1:1$,

$n(\text{acetanilide}) = 0.1\ \mathrm{mol}$

Therefore,

$m(\text{acetanilide}) = 0.1 \times 135 = 13.5\ \mathrm{g}$

6. Match with the form asked in the question

The mass is written as:

$\underline{\hspace{2cm}} \times 10^{-1}\ \mathrm{g}$

Now,

$13.5\ \mathrm{g} = 135 \times 10^{-1}\ \mathrm{g}$

So the required integer is:

$\boxed{135}$

7. Comparison with stored answer

Stored correct answer = $135$

Our derived answer also = $135$, so they agree.