It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is
[tex]S_4=4+\dfrac43+\dfrac49+\dfrac4{27}[/tex]
It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.
Multiply the sum by 1/3 and you have
[tex]\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}[/tex]
Now subtracting this from [tex]S_4[/tex] gives
[tex]S_4-\dfrac13S_4=4-\dfrac4{81}[/tex]
That is, all the matching terms will cancel. Now solving for [tex]S_4[/tex], you
have
[tex]\dfrac23S_4=4\left(1-\dfrac1{81}\right)[/tex]
[tex]S_4=6\left(1-\dfrac1{81}\right)[/tex]
[tex]S_4=\dfrac{480}{81}=\dfrac{160}{27}[/tex]
