Consider the above figure:

AD is the part of the ladder in the first case and AC is the ladder in the second case

AB = BD = 10 $\sqrt{7}$m          [given]

2AD = AC        [since AD is the half of the ladder AC]

$\therefore$ By Pythagoras Theorem in  $△$ABD,

${AD}^2$ =  ${AB}^2$ +  ${BD}^2$

⇒ ${AD}^2$ = $({10\sqrt{7})}^2$   +$({10\sqrt{7})}^2$ 
             = 2 $({10\sqrt{7})}^2$  

⇒$AD=10\sqrt{7}$$\sqrt{2}$

Since AD is half of the ladder, the complete length of the ladder

AC = 2AD = 20 $\sqrt{2}$$\sqrt{7}$

Now,
In  ΔABC,  
${AC}^2$ =  ${AB}^2$ +  ${BC}^2$

⇒ ${BC}^2={AC}^2-{AB}^2$
             $={\left(20\sqrt{7}\sqrt{2}\right)}^2$ -  $({10\sqrt{7})}^2$
             $=\left({10\sqrt{7})}^2\right[{\left(2\sqrt{2}\right)}^2-1]$
             = $({10\sqrt{7})}^2\times 7$

$\therefore BC=10\sqrt{7}\times \sqrt{7}=70m$