As I already noted in the comments, you should always and consistently use correct units. And also, the typography of the question itself was terrible but that’s not something you can influence. Anyway to get to answering the question …
I’m assuming that ‘grams of $\ce{MgBr2}$ t’ is supposed to mean something along the lines of theoretical yield. Also, I’m going to use $\mathrm{g/mol}$ as a unit for molar mass because that is what it should be. $\mathrm{amu}$ is a deprecated unit for atomic mass and cannot be used for molar mass without a conversion factor.
We thankfully are already given the reaction$$\ce{MgBr2 + Cl2 -> MgCl2 + Br2}\tag{1}$$ and it is thankfully the case that one mole of each reactant reacts to one mole of each product; i.e. all amounts of substance are theoretically equal. Thus, we can write the following:
$$\begin{align}n(\ce{MgBr2}) &= n (\ce{Br2})\tag{2.1}\\[0.4em]
M = \frac mn\tag{3} &\Longrightarrow n = \frac mM\\[0.6em]
\frac{m(\ce{MgBr2})}{M(\ce{MgBr2})} &= \frac{m(\ce{Br2})}{M(\ce{Br2})}\tag{2.2}\\[0.6em]
m(\ce{Br2}) &= \frac{m (\ce{MgBr2}) \times M(\ce{Br2})}{M(\ce{MgBr2})}\tag{2.3}\\[0.6em]
m(\ce{Br2}) &= \frac{176.6~\mathrm{g} \times 159.82~\mathrm{g/mol}}{184.13~\mathrm{g/mol}}\tag{2.4}\\[0.4em]
&= 153.28~\mathrm{g}\tag{2.5}\end{align}$$
Therefore, your calculation luckily turned out to be correct in spite of not using units.
The question also asks for a yield. Yields are typically given in percent obtained of the theoretical maximum. In the lab, I typically calculate yields from substance amounts but here we can use mass because we already have the data. When calculating a percentage, you need to remember to multiply by $100$ before adding the $\%$ sign — which could be interpreted as a shorthand for $1/100$. Thus:
$$\frac{135~\mathrm{g}}{153.28~\mathrm{g}}= 0.88 = 88~\%\tag{4}$$
The units should, naturally, always cancel out in yields.
I mentioned before that the question is ill-phrased. In a strict literal sense, the answer to ‘How many grams of $\ce{Br2}$ t are formed?’ — without any information on what the random ‘t’ is supposed to be — would indeed read $135~\mathrm{g}$ but unless you are in a quiz show that is unlikely to be the ‘correct’ answer.
