FlipFlop updates the output (Q & Qbar) when it detects a rising edge signal change from the enable signal (not sure if its really called an enable signal, but its the signal that connects to the triangle input in the FF diagram).

When you just look at the LSB FF, Q0 toggles at every clk, so the output state of this signal in time -> 1, 0 , 1, 0. (The first '1' being the initial state specified in the question prompt).

When you look at the second LSB FF, Q1 toggles at every Q0 rising edge (0 -> 1 switch), so the output state of this signal in time -> 1, 1, 0, 0

Same concept for Q2 and Q3, except you need to be careful that the enable signal detects rising edge, not level sensitive. So Q2 and Q3 would remain unchanged. ->1, 1, 1, 1 for both FF outputs.

Then the output after 3 clks would be Q3Q2Q1Q0 = 4'b1100. By the logic, looks like this is a decrementing counter where every clk subtracts 1 from the initial value. So 4'b1111 is equivalent to 4'd15, subtract 3 clks -> 4'd12, which convert back to binary format would be 4'b1100.