Strontium hydroxide or #SrOH_2# is a strong base and will completely ionize. This means that virtually all #SrOH_2# will dissociate in water into metal cations and hydroxide ions.
#"Sr"("OH")_2(s) -> "Sr"^{2+}+2"OH"^-#
#pOH# is the negative log of the concentration of hydroxide molecules. Since we know that #SrOH_2# will completely ionize in solution, we can use the given concentration to solve for #pOH#.
#pOH = -log [OH^-]#
#pOH = -log [2.0 * 10^-1 M]#
#pOH = 0.70#
The question asks for #pH# and not #pOH#. Using the formula below, we can solve for #pH#.
#pH + pOH = 14#
#pH + 0.70 = 14#
#pH = 14-0.70 = 13.3#
#pH = 13.3#
The #pH# is 13.3 at #"25 degrees Celsius"#