In the given circuit, the current flowing through the resistance...

Given, ${i}_{1} = 0 . 3A, {i}_{1} + {i}_{2} + {i}_{3} = 0 . 9A$

So, ${V}_{AB} = {i}_{1} \times 20Ω = 0 . 3 \times 20V = 6V$

${i}_{2} = \dfrac{6V}{15Ω} = \dfrac{2}{5}A$

Now,

${i}_{1} + {i}_{2} + {i}_{3} = \dfrac{9}{10}A$

$⇒ \dfrac{3}{10} + \dfrac{2}{5} + {i}_{3} = \dfrac{9}{10}$

$⇒ \dfrac{7}{10} + {i}_{3} = \dfrac{9}{10}$

$⇒ {i}_{3} = 0 . 2A$

So, ${i}_{3} \times {R}_{1} = 6V$

$⇒ ( 0 . 2 ) {R}_{1} = 6$

$⇒ {R}_{1} = \dfrac{6}{0 . 2} = 30Ω$