Answer
$a=10.3m/s^2$, directed at an angle of $21.9^o$ above the horizontal.
Work Step by Step
$m_{rocket}=4.5\times10^5kg$ and $F_{thrust}=7.5\times10^6N$, directed at $\theta=55^o$ above the horizontal.
In unit-vector notation, $$\vec{F}_{thrust}=(7.5\times10^6\cos55N)i+(7.5\times10^6\sin55N)j$$ $$\vec{F}_{thrust}=(4.3\times10^6N)i+(6.14\times10^6N)j$$ as we consider rightward and upward as $+x$ and $+y$ direction respectively.
Besides thrust force, gravitational force also acts upon the rocket in the downward $-y$ direction. $$m\vec{g}=-(4.5\times10^5\times9.8N)j$$ $$m\vec{g}=-(4.41\times10^6N)j$$
Therefore, $$\vec{F}_{net}=\vec{F}_{thrust}+m\vec{g}=(4.3\times10^6N)i+(1.73\times10^6N)j$$
Applying Newton's 2nd Law of Motion, $$\vec{a}=\frac{\vec{F}_{net}}{m_{rocket}}=\frac{(4.3\times10^6N)i+(1.73\times10^6N)j}{4.5\times10^5kg}$$ $$\vec{a}=(9.56m/s^{2})i+(3.84m/s^{2})j$$
Magnitude of $\vec{a}$: $$a=\sqrt{(9.56)^2+(3.84)^2}=10.3m/s^2$$
Let's call $\alpha$ the angle of $\vec{a}$ above the horizontal.
$$\tan\alpha=\frac{a_j}{a_i}=0.402$$ $$\alpha=21.9^o$$
