Answer: maximum length of the nanowire is 510 nm
Explanation:
   
From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K 
Thermal conductivity of silicon carbide k = 30 W/m.K 
Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m  
lets consider the equation for the value of m 
m = ( (hP/kAc)^1/2 )  = ( (4h/kD)^1/2 )  
m =  ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04    
now lets find the value of h/mk     
h/mk = 10⁵ / ( 942809.04 × 30) =  0.00353 
lets consider the value θ/θb by using the equation 
θ/θb = (T - T∞) / (T - T∞) 
θ/θb =  (3000 - 8000) / (2400 - 8000) 
= 0.893 
the temperature distribution at steady-state is expressed as;
θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)]   / [cosh mL+  (h/mk) sinh mL]
 θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)]   / [cosh mL+  (h/mk) sinh mL]
 θ/θb = [ 1 ]  / [cosh mL+  (h/mk) sinh mL]
so we substitute
0.893 =  [ 1 ]  / [cosh (942809.04 × L) +  (0.00353) sinh (942809.04 × L)]
L = 510 × 10⁻⁹m
L = 510 nm
therefore maximum length of the nanowire is 510 nm