$n_K=\dfrac{3,9}{39}=0,1mol \\PTHH : \\2K+2H_2O\to 2KOH+H_2 \\Theo\ pt : \\n_{KOH}=n_K=0,1mol \\⇒m_{KOH}=0,1.56=5,6g \\n_{H_2}=\dfrac{1}{2}.n_K=\dfrac{1}{2}.0,1=0,05mol \\⇒m_{H_2}=0,05.2=0,1g \\m_{dd\ spu}=3,9+100-0,1=103,8g \\⇒C\%_{KOH}=\dfrac{5,6}{103,8}.100\%=5,39\% $