The maximum allowed power dissipation for a 29.3-ω resistor is stated to be 10.0 w. find the largest current that this resistor can take safely without burning out.

0.584 amperes
   I am going to assume that 29.3-ω is a typo and actually means 29.3-Ω because of silly case folding.
   Using ohms law
  I = V/R
  IR=V
 And since power is voltage multiplied by amperage we have
P = IV
  Substitute IR for V
 P = I(IR)
  P = I^2*R
 P/R = I^2
 sqrt(P/R) = I
   Substitute the known values and calculate.
sqrt(10/29.3) = I
 sqrt(0.341296928) = I
 0.584206238 = I
   Rounding to 3 significant figures gives 0.584 amperes.

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