How do you solve this math problem differential equation using Laplace ?

Shouldn't it be L di/dt + R i = V ?

I'm going to assume that, because otherwise you don't even have a DIFFERENTIAL equation.  And also because I know that Ohms and Henrys are not the same units !

For now, though, let's dispense with the units.  You have

1 di/dt + 4 i = 6 sin(2t).

The Laplace transform of i we'll call F(s).

The Laplace transform of di/dt is s*F(s) - i(0), in your problem just s*F(s).

The Laplace transform of 6 sin(2t) is 12/(s^2 + 4).

So your differential equation becomes:

s*F(s) + 4*F(s) = 12/(s^2 + 4), or

F(s) = 12/[(s^2 + 4)(s + 4)].

Now you use partial fractions to pull it apart:

12/[(s^2 + 4)(s + 4)] = A/(s + 4) + (Bs + C)/(s^2 + 4) =>

12 = As^2 + 4A + Bs^2 + 4Bs + Cs + 4C =>

12 = 4A + 4C;

0 = 4B + C;

0 = A + B.

From the last two equations you get 0 = C - 4A, so the "12" equation becomes

12 = 4A + 16A => A = 3/5, B = -3/5, C = 12/5.

F(s) = (3/5)/(s+4) - (3/5)s/(s^2 + 4) + (12/5)/(s^2 + 4).

From a table of Laplace transforms you get

i(t) = (3/5)e^(-4t) - (3/5)cos(2t) + (6/5)sin(2t)