We can parameterize the curve by setting [tex]y=t[/tex], then
[tex]x^2=10t^3\implies x=\sqrt{10}\,t^{3/2}[/tex]
The arc length of [tex]x^2=10y^3[/tex], parameterized by [tex]\mathbf r(t)=(x(t),y(t))=(\sqrt{10}\,t^{3/2},t)[/tex], is given by the integral
[tex]\displaystyle\int_0^{10}\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt=\int_0^{10}\sqrt{\frac{45}2t+1}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac2{45}\int_1^{226}\sqrt u\,\mathrm du[/tex] where [tex]u=\dfrac{45}2t+1[/tex]
[tex]=\dfrac4{135}u^{3/2}\bigg|_1^{226}=\dfrac{4(226\sqrt{226}-1)}{135}\approx100.638[/tex]
