Start by determining the concentrations of the two species in the buffer. For acetic acid, you have
C=n/V⇒nacetic
=C⋅V
=0.2M⋅100⋅10^−3L
=20⋅10^−3moles
This means that the concentration of acetic acid in the buffer will be
Cacetic=nacetic/Vbuffer
=20⋅10^−3moles/(100+100)⋅10^- 3L
=0.1 M
For sodium acetate, the number of moles is
nacetate=C⋅V
=0.15xMx100⋅10^−3L
=15⋅10−3moles
As a result ,the concentration of MgOH in the buffer will be
CMgOH=nMgOH/Vbuffer
=10⋅10^−3moles/(100+50)x10^-3
= 0.6667
Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution
pHsolution=pKa+log([MgOH][CH3COOH])
pHsolution=4.75+log(0.0667M ÷0.1M)
=4.75−0.175=4.575