Explain:

Dear student,

A 6.90 M solution has 6.90 mol KOH in 1 L of solution.
Since , 1 L = 1000 cm³ , we can say that 
6.90 mol KOH is present in 1000 cm³ solution
Mass of 6.90 mol KOH = $6.90 \times Molar mass of KOH$
                                      = $6.90 \times 56 g mo{l}^{-1}= 386.4 g$

30% solution of KOH contains 30 g KOH in 100 g solution
1 g of KOH will be present in $\left(\frac{100}{30}\right) g$ solution
386.4 g of KOH will be present in$\left(\frac{100}{30}\right) \times 386.4 g$ solution = 1288 g solution 
Density of solution = $\left(\frac{Mass}{Volume}\right)$ = $\left(\frac{1288}{1000}\right)$ = 1.288 g/cm^-3
So, density of the solution = 1.288 g/cm^-3

Regards