Mole Concept

Let the masses of $\ce{H2O}$ and $\ce{NH3}$ in the solution be $m_{\ce{H2O}}$ and $m_{\ce{NH3}}$ respectively. Then, $m_{\ce{H2O}} = 105 \text{ g}$ and:

$\begin{aligned} \frac{m_{\ce{NH3}}}{m_{\ce{NH3}}+m_{\ce{H2O}}} & = 0.3 \\ m_{\ce{NH3}} & = 0.3m_{\ce{NH3}}+0.3m_{\ce{H2O}} \\ 0.7 m_{\ce{NH3}} & = 0.3 m_{\ce{H2O}} \\ \Rightarrow m_{\ce{NH3}} & = \frac{0.3 m_{\ce{H2O}}}{0.7} = \frac{0.3\times 105}{0.7} = 45 \text{ g} \end{aligned}$

The volume of the solution,

$\begin{aligned} A & = \frac{\text{mass of solution}}{\text{density of solution}} = \frac{m_{\ce{NH3}}+m_{\ce{H2O}}}{0.9} = \frac{45+105}{0.9} = 166.667 \text{ g} \end{aligned}$

The volume of $\text{45 g}$ of $\ce{NH3}$ at STP ( $0^\circ \text{C, 760 mm Hg}$ ):

$V_{STP} = \dfrac{45 \times 22.4}{\text{molar mass of }\ce{NH3}} = \dfrac{45 \times 22.4}{17} = \text{59.294 L}$

Using $\dfrac{pV}{T} = \text{constant}$ , the volume of $\text{45 g}$ of $\ce{NH3}$ at $4^\circ \text{C, 775 mm Hg}$ :

$\begin{aligned} \frac{775B}{277} & = \frac{760V_{STP}}{273} \\ \Rightarrow B & = \frac{277 \times 760 \times 59.294}{273 \times 775} = \text{58.998 L} \end{aligned}$

Therefore, $A+B = 166.667+ 58.998 = \boxed{225.67}$