Mol of $\ce{NaOH}$= $3/40$
Mol of $\ce{H2SO4} = 4.9/98.1 = 49/981$
If you compare the mol of both of them, you will see that the mol of $\ce{NaOH}$ is in excess.
You would now do $\ce{NaOH}$ - $\ce{H2SO4}$. By doing this you have used up all of the $\ce{H2SO4}$.
$3/40 - 49/981= 983/39240$
This is the mol of $\ce{Na2SO4}$. You can clearly see that the mol of $\ce{NaOH}$ is in excess.
Due to the molar ratio of $\ce{H2SO4}$= $\ce{Na2SO4}$ (1=1). The number of mol of $\ce{Na2SO4}$ made should be the amount of $\ce{H2SO4}$ which has all been used up.
Therefore, $49/981 = 0.0499 $
$0.0499 * 142.1 = 7.01 g $
I think answer given by OP is wrong.
Update:
More information given below.
Here is the equation with some information.

To make this explanation as easy as possible, the initial moles are only to 3 significant figures(So the actual answer at the end will not be accurate) and have been multiplied by 1000.
So you can see that $\ce{NaOH} $ makes up $75 mol $ and $\ce{H2SO4}$ is $55 mol $ . The "left in excess row " is the $\ce{2NaOH}$ -$\ce{H2SO4}$. You can easily see with theses numbers that there will be 20 mol of $\ce{2NaOH}$ left over and all of the $55 mol $ of $\ce{H2SO4}$ will be used up.
if $55 mol$ of $\ce{H2SO4}$ has been used up then 55 mol of $\ce{Na2SO4}$ has been produced, due to $1 to 1 $ molar ratio of those two molecules.
Then divide by 1000 as this was simply so the numbers are easier to look at.