${P_1} + \frac{1}{2}\rho v_1^2 = {P_2} + \frac{1}{2}\rho v_2^2\,\,\,...\left( i \right)$

From question,

${P_1} - {P_2} = 3 \times {10^5},\frac{{{A_1}}}{{{A_2}}} = 5$

According to equation of constinuity

${A_1}{v_1} = {A_2}{v_2}$

$or,\frac{{{A_1}}}{{{A_2}}} = \frac{{{v_2}}}{{{v_1}}} = 5$

$ \Rightarrow \,\,{v_2} = 5{v_1}$

From equation $(i)$

${P_1} - {P_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)$

$or\,\,3 \times {10^5} = \frac{1}{2} \times 1000\left( {5v_1^2 - v_1^2} \right.$

$ \Rightarrow 600 = 6{v_1} \times 4{v_1}$

$ \Rightarrow v_1^2 = 25$

$\therefore \,{v_1} = 5\,m/s$