Answer
a) $\approx 48 \;\rm cm$
b) $\approx 85\;\rm kg$
Work Step by Step
a) Since the two pistons are on the same level, we can use Pascal's law
$$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$$
where $A_1$ is the cross-sectional area of the student piston and $A_2$ is the cross-sectional area of the elephant piston.
$$\dfrac{m_{\rm student}\color{red}{\bf\not} g }{\color{red}{\bf\not} \pi r_1^2}=\dfrac{m_{\rm elephant }\color{red}{\bf\not} g}{\color{red}{\bf\not} \pi r_2^2}$$
Hence,
$$r_1^2=\dfrac{m_{\rm student}r_2^2}{m_{\rm elephant }}$$
The diameter is $D=2r$, so $r=D/2$.
$$\left(\dfrac{D_1}{2}\right)^2=\dfrac{m_{\rm student}r_2^2}{m_{\rm elephant }}$$
$$ {D_1} =2\left[\sqrt{\dfrac{m_{\rm student}r_2^2}{m_{\rm elephant }}}\right]$$
$$ {D_1} =2r_2 \left[\sqrt{\dfrac{m_{\rm student}}{m_{\rm elephant }}}\right]$$
Plugging the known;
$$ {D_1} =2 (1)\left[\sqrt{\dfrac{70 }{ 1200}}\right]$$
$$ {D_1}=\color{red}{\bf 0.483}\;\rm m$$
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b) Now when the two students stand on piston 1, it sinks 0.35 m.
So,
$$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}+h\rho_{oil}g$$
where $F_1=(m_{\rm student1}+m_{\rm student2})g$
$$\dfrac{(m_{\rm student1}+m_{\rm student2})\color{red}{\bf\not}g}{\pi r_1^2}=\dfrac{m_{\rm elephant }\color{red}{\bf\not}g}{\pi r_2^2}+h\rho_{oil}\color{red}{\bf\not}g$$
$$\dfrac{(m_{\rm student1}+m_{\rm student2}) }{\pi r_1^2} =\dfrac{m_{\rm elephant } }{\pi r_2^2}+h\rho_{oil} $$
$$ m_{\rm student1}+ m_{\rm student2} ={\pi r_1^2} \left[\dfrac{m_{\rm elephant } }{\pi r_2^2}+h\rho_{oil} \right]$$
$$ m_{\rm student2} ={\pi r_1^2} \left[\dfrac{m_{\rm elephant } }{\pi r_2^2}+h\rho_{oil} \right]- m_{\rm student1} $$
Plugging the known;
$$ m_{\rm student2} ={\pi \left(\frac{0.483}{2}\right)^2} \left[\dfrac{ 1200 }{\pi (1)^2}+(0.35)(900) \right]- 70 $$
$$ m_{\rm student2} =\color{red}{\bf 57.7}\;\rm kg$$
