Assume that the frustum is conical, because radii are implied.
Volume of large cone (enclosing the frustum) V1=⅓(πr12)h1; volume of small (top) cone V2=⅓(πr22)h2, where h=h1-h2, i.e., the difference of the heights of the two cones forming the frustum.
r1=63/11, r2=21/11, h=1490/121 units (given). [r1=3r2.]
Volume of frustum V=V1-V2=⅓(πr12)h1-⅓(πr22)h2=(π/3)(r12h1-r22h2).
h2/r2=h1/r1 because the two cones forming the frustum are similar figures.
h2=h1-h, so (h1-h)/r2=h1/r1,
r1(h1-h)=h1r2,
r1h1-r1h=h1r2,
h1(r1-r2)=r1h, h1=r1h/(r1-r2).
From the given info: h1=(63/11)(1490/121)/(42/11),
h1=(3/2)(1490/121)=2235/121.
h2=2235/121-1490/121=745/121. [h1=3h2.]
V=(π/3)(r12h1-r22h2)=745(π/3)(21/11)2[27/121)-1/121],
V=26(745π/3)(21/11)2/112=610.98 cu units approx.