Answer:
a)  Q = 397.57 pC
, Q = 3.18 104 pC
, b) C = 1.157 10⁻¹⁰ F
,  V = 3.4375 V
,
c)  U = 54.7 nJ
,  d) ΔU = 54 nJ, 
Explanation:
a) The capacity of a capacitor is defined
         C = Q / V
         Q = C V
          
can also be calculated using geometry consideration
         C = e or A / d
          
we reduce to the SI system
        A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²
        d = 1.53 cm = 1.53 10⁻² m
we substitute
          Q = eo A / d V
          Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275
          Q = 3.9757 10⁻¹⁰ C
          
let's reduce to pC
          Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)
           Q = 397.57 pC
when the capacitor is introduced into the water the dielectric constant is different
            Q = k Q₀
            Q = 80 397.57
            Q = 3.18 104 pC
b) Find capacitance and voltage after submerged in water
            C = k C₀
            C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²
            C = 1.157 10⁻¹⁰ F
            V = Vo / k
             V = 275/80
             V = 3.4375 V
c) The stored energy is
              U = ½ C V²
               U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²
              U = 5.47 10⁻⁸ J
let's reduce to nJ
               109 nJ = 1 J
                U = 54.7 nJ
d) energy after submerging
              U = ½ (kCo) (Vo / k) 2
              U = ½ Co Vo2 / k
              U = U₀ / k
              U = 54.7 / 80 nJ
               U = 0.68375 nJ
the energy change is
          ΔU = U₀ -U
           ΔU = 54.7 - 0.687375