Correct Answer - C

After first, `I_(1)=I_(0)/2`

After second, `I_(2)=I_(1) cos^(2) 30^(@)=I_(0)/2xx3/4=(3I_(0))/8`

After third, `I_(3)=I_(2) cos^(2) 30^(@)=(3I_(0))/8xx3/4=(9I_(0))/32`

After fourth, `I_(3) cos^(2) 30^(@)=(9I_(0))/32xx3/4=27/128I_(0)=0.21I_(0)`