India opener Rohit Sharma on Sunday became the third quickest batsman to 9000 ODI runs, achieving the feat against Australia during the third ODI at M. Chinnaswamy in Bengaluru .

Rohit took 217 innings to reach the milestone which is the third best after India skipper Virat Kohli (194) and AB De Villiers (205). The 32-year-old pipped greats like Sourav Ganguly (228) and Sachin Tendulkar (235).

Heading into the Bengaluru decider, he needed four runs to reach the milestone and did that in the first over of the Indian chase of 287.

Not long back, during the second ODI at Rajkot, the 2019 ODI cricketer of the year was the fastest to 7000 ODI runs as an opener. He had reached the milestone in merely 137 innings and surpassed Hashim Amla (147) and Sachin Tendulkar (160).

Sharma has been in ominous form in white-ball cricket since the ICC Cricket World Cup in England last year. During the CWC, he slammed five centuries before India were knocked out in the semi-finals.

In 2019, he slammed a total of seven centuries and amassed 1409 runs in 28 matches at an average of 57.30.

Meanwhile, in the Bengaluru decider, Australia won the toss and opted to bat first. Steve Smith struck a well-compiled century as Australia posted a competitive 286/9 in 50 overs.

Former skipper Smith brought up his ninth century in 117 balls, third against India and got out on 131 from 132 deliveries as he tried to change gears and was caught at deep midwicket by Shreyas Iyer off Mohammed Shami.