Let the resistance of the galvanometer coil be $R_g = 240\,\Omega$ and the resistance of the shunt be $R_s = 10\,\Omega.$

Since the coil and shunt are connected in parallel, the potential difference across both is the same.

If current through the coil is $I_g$ and through the shunt is $I_s$, then $I_g R_g = I_s R_s.$ So, $\frac{I_s}{I_g} = \frac{R_g}{R_s} = \frac{240}{10} = 24.$

Hence total current through the ammeter is $I = I_g + I_s = I_g + 24I_g = 25I_g.$

From the figure, the current through the coil is $6.4\,\text{mA}$. Therefore, $I = 25 \times 6.4\,\text{mA} = 160\,\text{mA}.$

Therefore the ammeter reading is $\boxed{160\,\text{mA}}.$